Solution: the δhvap of a certain compound is 48.17 kj mol-1 and its δsvap is 52.91 j mol-1•k-1. what is the boiling point of this compound?

910.44K

Explanation:

From the question, the parameters given are; ∆H(vaporization)= 48.17 kJ/mol and ΔS(vaporization)= 52.91 J mol-1•K.

Using the formula;

dG = dH - TdS (1).

Here, dG=0, dG= change in free energy.

dH(vaporization) = 48,170 J/mol

dS(vaporization) = 52.91 J/mol.K

Boiling point of the substance, T=???

Therefore, we solve for T(temperature, that is the boiling point-boiling temperature) of the compound in Kelvin.

Slotting in the parameters given into equation (1). We have;

0= 48,170 J/mol - T× 52.91 J/mol.K

=> 48,170= 52.91T

T= 910.44 K